Question

A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.​

Answer 1

Initial velocity of the stone, u = 0

Final velocity of the stone, v = ?

Height of the tower, h = 19.6 m

Acceleration due to gravity, g = 9.8 ms-2

For a freely falling body:

v2 - u2 = 2 g h

v2 - 02 = 2 × 9.8 × 19.6

v2 = 19.6 × 19.6 = (19.6)2 

v2 = 384.16



v = 19.6 ms-1

Hence, the velocity of the stone just before touching the ground is 19.6 ms-1.

Concept Note: When a body is falling vertically downwards, its velocity increases, so the acceleration due to gravity g is taken as positive.

Answer 2

_/\_Hello mate__here is your answer--

u = 0 m/s

v = ?

s = Height of the stone = 19.6 m

g = 9.8 ms−2

According to the equation of motion under gravity

v^2−u^2=2gs

⇒ v^2−(0)^2=2×9.8×19.6

⇒ v^2=2×9.8 ×19.6=(19.6)^2

⇒ v=19.6 ms−1

Hence, the velocity of the stone just before touching the ground is 19.6 ms^−1.

I hope, this will help you.☺

Thank you______❤

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